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{{definitioncard|Uniformly integrable| | |||
Let <math>(\Omega,\F,(\F_n)_{n\geq0},\p)</math> be a filtered probability space. A family <math>(X_i)_{i\in I}</math> of r.v.'s in <math>L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)</math>, indexed by an arbitrary index set <math>I</math>, is called uniformly integrable (denoted by u.i.) if | |||
<math display="block"> | |||
\lim_{a\to\infty}\sup_{i\in I}\E[\vert X_i\vert \one_{\{\vert X_i\vert > a\}}]=0. | |||
</math> | |||
}} | |||
{{alert-info | | |||
A singel r.v. in <math>L^1(\Omega,\F,(\F_n)_{n\geq0},\p)</math> is always u.i. (this follows from dominated convergence). If <math>\vert I\vert < \infty</math>, then using | |||
<math display="block"> | |||
\vert X_i\vert \leq \sum_{j\in I}\vert X_j\vert, | |||
</math> | |||
we get | |||
<math display="block"> | |||
\E[\vert X_i\vert \one_{\{\vert X_i\vert > a\}}]\leq \sum_{j\in I}\E[\vert X_j\vert\one_{\{ X_j\vert > a\}}]\xrightarrow{a\to\infty}0. | |||
</math> | |||
Let <math>(X_i)_{i\in I}</math> be u.i. For <math>a</math> large enough, we then have | |||
<math display="block"> | |||
\sup_{i\in I}\E[\vert X_i\vert \one_{\{\vert X_i\vert > a\}}]\leq 1. | |||
</math> | |||
Hence | |||
<math display="block"> | |||
\sup_{i\in I}\E[\vert X_i\vert]=\sup_{i\in I}\E[\vert X_i\vert (\one_{\{\vert X_i\vert\leq a\}}+\one_{\{ \vert X_i\vert > a\}})]\leq 1+a < \infty, | |||
</math> | |||
which implies that <math>(X_i)_{i\in I}</math> is bounded in <math>L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)</math>. | |||
}} | |||
'''Example''' | |||
Let <math>Z\in L^1(\Omega,\F,(\F_n)_{n\geq0},\p)</math>. Then the family | |||
<math display="block"> | |||
\Theta=\{X\in L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)\mid \vert X\vert\leq Z\} | |||
</math> | |||
is u.i. Indeed, we have | |||
<math display="block"> | |||
\sup_{X\in\Theta}\E[\vert X\vert \one_{\{\vert X\vert > a\}}]\leq \E[Z\one_{\{Z > a\}}]\xrightarrow{a\to\infty}0. | |||
</math> | |||
'''Example''' | |||
Let <math>\phi:\R_+\to\R_+</math> be a measurable map, such that | |||
<math display="block"> | |||
\frac{\phi(x)}{x}\xrightarrow{x\to\infty}\infty. | |||
</math> | |||
Then for all <math>C > 0</math>, the family | |||
<math display="block"> | |||
\Theta_C=\{X\in L^1(\Omega,\F,(\F_n)_{n\geq0},\p)\mid \E[\phi(\vert X\vert)]\leq C\} | |||
</math> | |||
is u.i. Indeed, for <math>a</math> large enough, we have | |||
<math display="block"> | |||
\E[\vert X\vert \one_{\{\vert X\vert > a\}}]=\E\left[\frac{\vert X\vert}{\phi(\vert X\vert)}\phi(\vert X\vert)\one_{\{\vert X\vert > a\}}\right]\leq \sup_{x > a}\left(\frac{x}{\phi(x)}\right)\E[\phi(\vert X\vert)]\leq C\sup_{x > a}\left(\frac{x}{\phi(x)}\right)\xrightarrow{a\to\infty}0. | |||
</math> | |||
Thus | |||
<math display="block"> | |||
\sup_{X\in\Theta}\E[\vert X\vert\one_{\{\vert X\vert > a\}}]\leq C\sup_{x > a}\left(\frac{x}{\phi(x)}\right) | |||
</math> | |||
{{proofcard|Proposition|prop-1|Let <math>(\Omega,\F,(\F_n)_{n\geq0},\p)</math> be a filtered probability space. Let <math>(X_i)_{i\in I}</math> be a family of r.v.'s bounded in <math>L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)</math>, i.e. <math>\sup_{i\in I}\E[\vert X_i\vert] < \infty</math>. Then <math>(X_i)_{i\in I}</math> is u.i. if and only if for all <math>\varepsilon > 0</math> there is a <math>\delta > 0</math> such that for all <math>\A\in \F</math>, if <math>\p[A] < \delta</math> then <math>\sup_{i\in I}\E[\vert X_i\vert \one_A] < \varepsilon</math>. | |||
|For the right implication, let <math>\varepsilon > 0</math>. Then there exists <math>a > 0</math> such that | |||
<math display="block"> | |||
\sup_{i\in I}\E[\vert X_i\vert \one_{\{\vert X_i\vert > a\}}] < \frac{\varepsilon}{2}. | |||
</math> | |||
Now let <math>\delta=\frac{\varepsilon}{2a}</math> and <math>A\in\F</math> such that <math>\p[A] < \delta</math>. Then | |||
<math display="block"> | |||
\E[\vert X_i\vert \one_A]=\E[\vert X_i\vert \one_A\one_{\{\vert X_i\vert > a\}}]+\E[\vert X_i\vert \one_A\one_{\vert X_i\vert \leq a\}}]\leq \E[\vert X_i\vert \one_{\{\vert X_i\vert > a\}}]+a\underbrace{\E[\one_A]}_{\p[A]} < \frac{\varepsilon}{2}+a\delta < \frac{\varepsilon}{2}+a\frac{\varepsilon}{2a}=\varepsilon. | |||
</math> | |||
For the left implication, let <math>C=\sup_{i\in I}\E[\vert X_i\vert] < \infty</math>. From Markov's inequality,we get | |||
<math display="block"> | |||
\p[\vert X_i\vert > a]\leq \frac{\E[\vert X_i\vert]}{a}\leq \frac{C}{a}. | |||
</math> | |||
Now let <math>\delta > 0</math> such that <math>(ii)</math> holds. If <math>\frac{C}{a} < \delta</math>, then for all <math>i\in I</math> | |||
<math display="block"> | |||
\E[\vert X_i\vert \one_{\{\vert X_i\vert > a\}}] < \varepsilon. | |||
</math>}} | |||
{{proofcard|Corollary|cor-1|Let <math>(\Omega,\F,(\F_n)_{n\geq0},\p)</math> be a filtered probability space. Let <math>X</math> be a bounded r.v., i.e. <math>X\in L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)</math>. Then the family | |||
<math display="block"> | |||
\{\E[X\mid\mathcal{G}]\mid \mathcal{G}\subset \F,\text{ $\mathcal{G}$ is a $\sigma$-Algebra}\} | |||
</math> | |||
is u.i. | |||
|Let <math>\varepsilon > 0</math>. Then there exists a <math>\delta > 0</math> such that for all <math>A\in \F</math>, if <math>\p[A] < \delta</math> then <math>\E[\vert X\vert \one_A] < \varepsilon</math>. Then for all <math>a > 0</math>, we have | |||
<math display="block"> | |||
\p[\E[X\mid \mathcal{G}] > a]\leq \frac{\E[\vert \E[X\mid \mathcal{G}]\vert]}{a}\leq \frac{\E[\vert X\vert]}{a}. | |||
</math> | |||
For <math>a</math> large enough, i.e. <math>\frac{\E[\vert X\vert]}{a} < \delta</math>, we have | |||
<math display="block"> | |||
\E[\vert \E[X\mid \mathcal{G}]\vert \one_{\{ \vert \E[X\mid \mathcal{G}]\vert > a\}}]\leq \E[\E[\vert X\vert \one_{\vert \E[X\mid \mathcal{G}]\vert > a\}}\mid \mathcal{G}]]\leq \E[\vert X\vert \one_{\{\E[X\mid \mathcal{G}]\vert > a\}}] < \varepsilon. | |||
</math>}} | |||
{{proofcard|Theorem|thm-1|Let <math>(\Omega,\F,(\F_n)_{n\geq0},\p)</math> be a filtered probability space. Let <math>(X_n)_{n\geq 0}</math> be a sequence of r.v.'s in <math>L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)</math>, which converges in probability to <math>X_\infty</math>. Then <math>X_n\xrightarrow{n\to\infty\atop L^1}X_\infty</math> if and only if <math>(X_n)_{n\geq 0}</math> is u.i. | |||
|For the right implication, we first note that <math>(X_n)_{n\geq 0}</math> is bounded in <math>L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)</math> since it converges in <math>L^1</math>. For <math>\varepsilon > 0</math>, there exists <math>N\in\N</math> such that for <math>n\geq N</math> we get | |||
<math display="block"> | |||
\E[\vert X_N-X_n\vert] < \frac{\varepsilon}{2}. | |||
</math> | |||
Next we note that <math>\{X_0,X_1,...,X_N\}</math> is u.i. since it is a finite family of bounded r.v.'s. Therefore, there exists a <math>\delta > 0</math>, such that for all <math>A\in\F</math>, if <math>\p[A] < \delta</math> then <math>\E[\vert X_n\vert\one_A] < \frac{\varepsilon}{2}</math> for all <math>n\in\{0,1,...,N\}</math>. Finally for <math>n\geq N</math>, we get | |||
<math display="block"> | |||
\E[\vert X_n\vert \one_A]\leq \E[\vert X_N-X_n\vert \one_A]+\E[\vert X_N\vert \one_A] < \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. | |||
</math> | |||
Thus <math>(X_n)_{n\geq 0}</math> is u.i. | |||
For the left implication, we note that if <math>(X_n)_{n\geq 0}</math> is u.i., then the family <math>(X_n-X_m)_{(n,m)\in\N^2}</math> is also u.i. since | |||
<math display="block"> | |||
\E[\vert X_n-X_m\vert \one_A]\leq \E[\vert X_n\vert \one_A]+\E[X_m\vert \one_A] | |||
</math> | |||
for all <math>A\in\F</math>. Now for <math>\varepsilon > 0</math>, there exists <math>a</math> sufficiently large, such that | |||
<math display="block"> | |||
\E[\vert X_n-X_m\vert \one_{\{\vert X_n-X_m\vert > a\}}] < \varepsilon. | |||
</math> | |||
Moreover, we note that | |||
<math display="block"> | |||
\E[\vert X_m-X_m\vert]\leq \E[\vert X_n-X_m\vert \one_{\{ \vert X_n-X_m\vert < \varepsilon\}}]+\E[\vert X_n-X_m\vert \one_{\{\vert X_n-X_m\vert\geq \varepsilon\}}] | |||
</math> | |||
<math display="block"> | |||
\leq \varepsilon +\E[\vert X_n-X_m\vert\one_{\{\vert X_n-X_m\vert \geq\varepsilon\}}\one_{\{\vert X_n-X_m\vert \leq a\}}]+\E[\vert X_n-X_m\vert\one_{\{\vert X_n-X_m\vert\geq \varepsilon\}}\one_{\{\vert X_n-X_m\vert > a\}}] | |||
</math> | |||
<math display="block"> | |||
\leq \varepsilon+a\p[\vert X_n-X_m\vert\geq\varepsilon]+\E[\vert X_n-X_m\vert \one_{\{\vert X_n-X_m\vert > a\}}]. | |||
</math> | |||
Then, using that <math>\lim_{n\to\infty}\p[\vert X_n-X_m\vert\geq\varepsilon]=0</math>, we can show that the right hand side converges to zero for <math>a</math> large enough, which implies that <math>(X_n)_{n\geq 0}</math> is a Cauchy sequence <math>L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)</math> and hence converges in <math>L^1</math>.}} | |||
Combining all our previous results, we have; if <math>(X_n)_{n\geq 0}</math> is a martingale, then the following are equivalent | |||
<ul style{{=}}"list-style-type:lower-roman"><li><math>(X_n)_{n\geq 0}</math> converges a.s. and in <math>L^1</math>. | |||
</li> | |||
<li><math>(X_n)_{n\geq 0}</math> is u.i. | |||
</li> | |||
<li><math>(X_n)_{n\geq 0}</math> is regular and <math>X_n=\E[X_\infty\mid \F_n]</math> a.s. | |||
</li> | |||
</ul> | |||
==General references== | |||
{{cite arXiv|last=Moshayedi|first=Nima|year=2020|title=Lectures on Probability Theory|eprint=2010.16280|class=math.PR}} |
Latest revision as of 01:53, 8 May 2024
Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. A family [math](X_i)_{i\in I}[/math] of r.v.'s in [math]L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)[/math], indexed by an arbitrary index set [math]I[/math], is called uniformly integrable (denoted by u.i.) if
A singel r.v. in [math]L^1(\Omega,\F,(\F_n)_{n\geq0},\p)[/math] is always u.i. (this follows from dominated convergence). If [math]\vert I\vert \lt \infty[/math], then using
Example
Let [math]Z\in L^1(\Omega,\F,(\F_n)_{n\geq0},\p)[/math]. Then the family
is u.i. Indeed, we have
Example
Let [math]\phi:\R_+\to\R_+[/math] be a measurable map, such that
Then for all [math]C \gt 0[/math], the family
is u.i. Indeed, for [math]a[/math] large enough, we have
Thus
Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. Let [math](X_i)_{i\in I}[/math] be a family of r.v.'s bounded in [math]L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)[/math], i.e. [math]\sup_{i\in I}\E[\vert X_i\vert] \lt \infty[/math]. Then [math](X_i)_{i\in I}[/math] is u.i. if and only if for all [math]\varepsilon \gt 0[/math] there is a [math]\delta \gt 0[/math] such that for all [math]\A\in \F[/math], if [math]\p[A] \lt \delta[/math] then [math]\sup_{i\in I}\E[\vert X_i\vert \one_A] \lt \varepsilon[/math].
For the right implication, let [math]\varepsilon \gt 0[/math]. Then there exists [math]a \gt 0[/math] such that
Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. Let [math]X[/math] be a bounded r.v., i.e. [math]X\in L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)[/math]. Then the family
Let [math]\varepsilon \gt 0[/math]. Then there exists a [math]\delta \gt 0[/math] such that for all [math]A\in \F[/math], if [math]\p[A] \lt \delta[/math] then [math]\E[\vert X\vert \one_A] \lt \varepsilon[/math]. Then for all [math]a \gt 0[/math], we have
Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. Let [math](X_n)_{n\geq 0}[/math] be a sequence of r.v.'s in [math]L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)[/math], which converges in probability to [math]X_\infty[/math]. Then [math]X_n\xrightarrow{n\to\infty\atop L^1}X_\infty[/math] if and only if [math](X_n)_{n\geq 0}[/math] is u.i.
For the right implication, we first note that [math](X_n)_{n\geq 0}[/math] is bounded in [math]L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)[/math] since it converges in [math]L^1[/math]. For [math]\varepsilon \gt 0[/math], there exists [math]N\in\N[/math] such that for [math]n\geq N[/math] we get
Next we note that [math]\{X_0,X_1,...,X_N\}[/math] is u.i. since it is a finite family of bounded r.v.'s. Therefore, there exists a [math]\delta \gt 0[/math], such that for all [math]A\in\F[/math], if [math]\p[A] \lt \delta[/math] then [math]\E[\vert X_n\vert\one_A] \lt \frac{\varepsilon}{2}[/math] for all [math]n\in\{0,1,...,N\}[/math]. Finally for [math]n\geq N[/math], we get
For the left implication, we note that if [math](X_n)_{n\geq 0}[/math] is u.i., then the family [math](X_n-X_m)_{(n,m)\in\N^2}[/math] is also u.i. since
Combining all our previous results, we have; if [math](X_n)_{n\geq 0}[/math] is a martingale, then the following are equivalent
- [math](X_n)_{n\geq 0}[/math] converges a.s. and in [math]L^1[/math].
- [math](X_n)_{n\geq 0}[/math] is u.i.
- [math](X_n)_{n\geq 0}[/math] is regular and [math]X_n=\E[X_\infty\mid \F_n][/math] a.s.
General references
Moshayedi, Nima (2020). "Lectures on Probability Theory". arXiv:2010.16280 [math.PR].